Question

Prove the following identity

Answer 1

hey mate,
here is the answer
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LHS =

$= >( \frac{ \cos(\theta) - \sin(\theta) + 1 }{ \cos(\theta) + \sin(\theta) - 1 } )$

$\frac{ (\cos(\theta) + 1) - \sin(\theta) }{ (\cos(\theta) + \sin(\theta)) - 1 } \times \frac{ (\cos(\theta) + 1) \sin(\theta) }{(\cos(\theta) + \sin(\theta)) + 1}$

$= > ( \frac{ (\cos(\theta) + 1) {}^{2} - \sin {}^{2} (\theta) }{( \cos(\theta) + \sin(\theta)) {}^{2} - {1}^{2} } )$

$= > \frac{\cos {}^{2}(\theta) + 1 + 2 \cos(\theta) - \sin {}^{2}(\theta) }{ \cos {}^{2}(\theta) + \sin {}^{2} (\theta) + 2 \sin(\theta)\cos(\theta) - 1 }$

$= > \frac{ \cos {}^{2} (\theta) + 2 \cos(\theta) + \cos {}^{2} (\theta) }{1 + 2 \sin(\theta) \cos(\theta) - 1}$

$= > ( \frac{ 2 \cos(\theta) + 2 \cos {}^{2} (\theta)}{2 \sin(\theta) \cos(\theta) })$

$= > (\frac{2 \cos(\theta)(1 + \cos(\theta)) }{2 \sin(\theta) \cos(\theta) } )$

$= > (\frac{1 + \cos(\theta) }{ \sin(\theta) } )$

$= > \frac{1}{ \sin(\theta) } + \frac{ \cos(\theta) }{ \sin(\theta ) }$

$= > \cosec(\theta) + \cot(\theta)$

RHS
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Answer 2

Hope this will help you